Physical Chemistry for Energy Engineering (20

for Energy Engineering (20 ^th : 2018/11/26)

Takuji Oda

Associate Professor, Department of Nuclear Engineering Seoul National University

*The class follows the text book: D.A. McQuarrie, J.D. Simon, “Physical Chemistry: A Molecular Approach", University Science Books (1997).

Course schedule (as of Nov. 5)

29-Oct No lecture

31-Oct

14 5-Nov 2. Phase equilibrium-1 15 7-Nov 2. Phase equilibrium-2 16 12-Nov 3. Chemical equilibrium-1 17 14-Nov Answers of homework-2 18 19-Nov Exam-02 (2 hour)

19 21-Nov 3. Chemical equilibrium-2 20 26-Nov 3. Chemical equilibrium-3 21 28-Nov 3. Chemical equilibrium-4 22 3-Dec 3. Chemical kinetics-1 23 5-Dec 3. Chemical kinetics-2 24 10-Dec 3. Chemical kinetics-3 25 12-Dec Answers of homework-3 26 17-Dec Exam-03 (2 hour)

(Example-6) Consider “ H₂ (g) + ½ O₂ (g) ⇌ H2O (l) ”, for which ∆_𝑟𝑟𝐺𝐺𝐺 = −237 kJ 𝑚𝑚𝑚𝑚𝑚𝑚⁻¹ at 298.15 K.

In this case, ∆_𝑟𝑟𝐺𝐺𝐺 has a large negative value, thus basically H₂O (l) is much more stable than the reactants at 298.15 K. However, a mixture of H₂ (g) and O₂ (g) remains unchanged.

If a spark or a catalyst is introduced, then the reaction occurs explosively.

 The “no” is emphatic in thermodynamics: If thermodynamics insists that a certain process will not occur spontaneously, then it will not occur.

 On the other hand, the “yes” is actually “maybe”. The fact that a process will occur spontaneously does not imply that it will occur at a detectable rate.

 Diamond remains its form, although a graphite is more favorable energetically, is another example.

The speed of reaction can be analyzed in the framework of rate theory. (next topic)

3.1. Chemical equilibrium for gases

- $26-6: The sign of _{∆𝑟𝑟𝐺𝐺} and not that of _{∆𝑟𝑟𝐺𝐺𝐺 𝑇𝑇} determines the direction of reaction spontaneity -

Theory of equilibrium is powerful, but it cannot indicate how long it take to reach the equilibrium state.

Kinetics and rate theory can analyze and give information on the speed of a reaction.

Index

time Spontaneous

process

Equilibrium

Kinetics and rate theory are often based on thermodynamics, but sometimes extended to microscopic systems (via statistics mechanics/thermodynamics).

Sufficiently long time

Insufficient Kinetics /

rate theory

We utilize the Gibbs-Helmholtz equation:

Substitute ∆_𝑟𝑟𝐺𝐺𝐺 = −𝑅𝑅𝑇𝑇 ln 𝐾𝐾_𝑃𝑃 (note this is a definition of ∆_𝑟𝑟𝐺𝐺𝐺, not a condition achieved at equilibrium state) for this equation:

𝜕𝜕∆𝐺𝐺𝐺 𝑇𝑇⁄

𝜕𝜕𝑇𝑇 _𝑃𝑃 = − ∆𝐻𝐻𝐺 𝑇𝑇²

𝜕𝜕 ln 𝐾𝐾_𝑃𝑃(𝑇𝑇)

𝜕𝜕𝑇𝑇 _𝑃𝑃 = 𝑑𝑑 ln 𝐾𝐾_𝑃𝑃(𝑇𝑇)

𝑑𝑑𝑇𝑇 = ∆_𝑟𝑟𝐻𝐻𝐺 𝑅𝑅𝑇𝑇² This means that

 If ∆_𝑟𝑟𝐻𝐻𝐺 > 0 (endothermic reaction), 𝐾𝐾_𝑃𝑃(𝑇𝑇) increases with temperature.

 If ∆_𝑟𝑟𝐻𝐻𝐺 < 0 (exothermic reaction), 𝐾𝐾_𝑃𝑃(𝑇𝑇) decreases with temperature.

Integrate the equation: ln𝐾𝐾_𝑃𝑃(𝑇𝑇₂)

𝐾𝐾_𝑃𝑃(𝑇𝑇₁) = �_𝑇𝑇1

𝑇𝑇₂ ∆_𝑟𝑟𝐻𝐻𝐺(𝑇𝑇) 𝑅𝑅𝑇𝑇² 𝑑𝑑𝑇𝑇

If the temperature range is small enough to consider ∆_𝑟𝑟𝐻𝐻𝐺 constant:

ln𝐾𝐾_𝑃𝑃(𝑇𝑇₂)

𝐾𝐾_𝑃𝑃(𝑇𝑇₁) = −

∆_𝑟𝑟𝐻𝐻𝐺 𝑅𝑅

1

𝑇𝑇₂ − 1 𝑇𝑇₁

3.1. Chemical equilibrium for gases

- $26-7: The variation of an equilibrium constant with temperature is given by the Van’t Hoff Equation -

(Example-7) Consider “ PCl₃(g) + Cl₂ (g) ⇌ PCl2 (g) ”.

Given that ∆_𝑟𝑟𝐻𝐻𝐺 has an average value of -69.8 kJ mol^-1 over 500-700 K and 𝐾𝐾_𝑃𝑃 is 0.0408 at 500K, evaluate 𝐾𝐾_𝑃𝑃 at 700K.

As ∆_𝑟𝑟𝐻𝐻𝐺 can be assumed as a constant over the concerned temperatures, ln𝐾𝐾_𝑃𝑃(𝑇𝑇₂)

𝐾𝐾_𝑃𝑃(𝑇𝑇₁) = −

∆_𝑟𝑟𝐻𝐻𝐺 𝑅𝑅

1

𝑇𝑇₂ − 1 𝑇𝑇₁ Substituting provided values gives:

ln𝐾𝐾_𝑃𝑃(700 𝐾𝐾)

𝐾𝐾_𝑃𝑃(500 𝐾𝐾) = ln

𝐾𝐾_𝑃𝑃(700 𝐾𝐾)

0.0408 = −

−69.8 × 10³ 𝑅𝑅

1 700 −

1 500 𝐾𝐾_𝑃𝑃 700 𝐾𝐾 = 3.36 × 10⁻⁴

Note that since the reaction is exothermic, 𝐾𝐾_𝑃𝑃 700 𝐾𝐾 is less than 𝐾𝐾_𝑃𝑃 500 𝐾𝐾 ; namely less product (PCl₂) at higher temperatures as 𝐾𝐾_𝑃𝑃 = ^{𝑃𝑃𝑃𝑃𝐶𝐶𝐶𝐶2}�_{𝑃𝑃𝑃𝑃𝐶𝐶𝐶𝐶3𝑃𝑃𝐶𝐶𝐶𝐶2} .

ln𝐾𝐾_𝑃𝑃(𝑇𝑇₂)

𝐾𝐾_𝑃𝑃(𝑇𝑇₁) = �_𝑇𝑇1

𝑇𝑇₂∆_𝑟𝑟𝐻𝐻𝐺(𝑇𝑇) 𝑅𝑅𝑇𝑇² 𝑑𝑑𝑇𝑇

Finally, we consider the temperature dependence of ∆_𝑟𝑟𝐻𝐻𝐺 constant in:

This equation can write down as:

ln 𝐾𝐾_𝑃𝑃(𝑇𝑇) = ln 𝐾𝐾_𝑃𝑃(𝑇𝑇₁) + �

𝑇𝑇₁

𝑇𝑇 ∆_𝑟𝑟𝐻𝐻𝐺(𝑇𝑇𝑇) 𝑅𝑅𝑇𝑇𝑇² 𝑑𝑑𝑇𝑇𝑇

For example, as we learned, the temperature dependence of ∆_𝑟𝑟𝐻𝐻𝐺 may be written as:

∆_𝑟𝑟𝐻𝐻𝐺 𝑇𝑇₂ = ∆_𝑟𝑟𝐻𝐻𝐺 𝑇𝑇₁ + �

𝑇𝑇₁ 𝑇𝑇₂

∆𝐶𝐶𝐺_𝑃𝑃 𝑇𝑇 𝑑𝑑𝑇𝑇 or expanded in respect to temperature as:

∆_𝑟𝑟𝐻𝐻𝐺 𝑇𝑇 = 𝛼𝛼 + 𝛽𝛽𝑇𝑇 + 𝛾𝛾𝑇𝑇² + 𝛿𝛿𝑇𝑇³ + ⋯ In this latter case, ln 𝐾𝐾_𝑃𝑃(𝑇𝑇) becomes:

ln 𝐾𝐾_𝑃𝑃(𝑇𝑇) = − 𝛼𝛼 𝑅𝑅𝑇𝑇 +

𝛽𝛽

𝑅𝑅 ln 𝑇𝑇 + 𝛾𝛾

𝑅𝑅 𝑇𝑇 + 𝛿𝛿

2𝑅𝑅 𝑇𝑇² + (𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑚𝑚𝑖𝑖_𝑐𝑐𝑚𝑚𝑖𝑖𝑐𝑐𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖) …

3.1. Chemical equilibrium for gases

- $26-7: The variation of an equilibrium constant with temperature is given by the Van’t Hoff Equation -

ln 𝐾𝐾_𝑃𝑃(𝑇𝑇) = ln 𝐾𝐾_𝑃𝑃(𝑇𝑇₁) + �

𝑇𝑇₁

𝑇𝑇 ∆_𝑟𝑟𝐻𝐻𝐺(𝑇𝑇𝑇) 𝑅𝑅𝑇𝑇𝑇² 𝑑𝑑𝑇𝑇𝑇

∆_𝑟𝑟𝐻𝐻𝐺 𝑇𝑇₂ = ∆_𝑟𝑟𝐻𝐻𝐺 𝑇𝑇₁ + �

𝑇𝑇₁ 𝑇𝑇₂

∆𝐶𝐶𝐺_𝑃𝑃 𝑇𝑇 𝑑𝑑𝑇𝑇

∆𝐶𝐶_𝑃𝑃𝐺 = 𝐶𝐶_𝑃𝑃𝐺 𝑁𝑁𝐻𝐻₃ 𝑖𝑖 − �1

2 𝐶𝐶^𝑃𝑃𝐺 𝑁𝑁₂ 𝑖𝑖 − �3

2 𝐶𝐶^𝑃𝑃𝐺 𝐻𝐻₂ 𝑖𝑖 Thus

ln 𝐾𝐾_𝑃𝑃(𝑇𝑇) = ln 𝐾𝐾_𝑃𝑃(725 𝐾𝐾) + �

725

𝑇𝑇 ∆_𝑟𝑟𝐻𝐻𝐺(𝑇𝑇𝑇)

𝑅𝑅𝑇𝑇^′2 𝑑𝑑𝑇𝑇^′

= 12.06 + 4583 𝑇𝑇 − 3.749 ln 𝑇𝑇 + 1.857 × 10⁄ ⁻³𝑇𝑇 − 0.118 × 10⁻⁶𝑇𝑇² (Example-8) Consider “ ½ N₂ (g) + 3/2 H₂ (g) ⇌ NH3 (g) ”.

The molar heat capacities of these species are give over 300-1500K as:

𝐶𝐶_𝑃𝑃𝐺 𝑁𝑁₂ 𝑖𝑖 𝐽𝐽 𝐾𝐾⁄ ⁻¹ 𝑚𝑚𝑚𝑚𝑚𝑚⁻¹ = 24.98 + 5.912 × 10⁻³𝑇𝑇 − 0.03376 × 10⁻⁶𝑇𝑇² 𝐶𝐶_𝑃𝑃𝐺 𝐻𝐻₂ 𝑖𝑖 𝐽𝐽 𝐾𝐾⁄ ⁻¹ 𝑚𝑚𝑚𝑚𝑚𝑚⁻¹ = 29.07 − 0.8368 × 10⁻³𝑇𝑇 + 2.012 × 10⁻⁶𝑇𝑇² 𝐶𝐶_𝑃𝑃𝐺 𝑁𝑁𝐻𝐻₃ 𝑖𝑖 𝐽𝐽 𝐾𝐾⁄ ⁻¹ 𝑚𝑚𝑚𝑚𝑚𝑚⁻¹ = 25.93 + 32.58 × 10⁻³𝑇𝑇 − 3.046 × 10⁻⁶𝑇𝑇²

Given that ∆_𝑓𝑓𝐻𝐻𝐺 𝑁𝑁𝐻𝐻₃(𝑖𝑖) = −46.11 𝑘𝑘𝐽𝐽 𝑚𝑚𝑚𝑚𝑚𝑚⁻¹𝑖𝑖𝑖𝑖 300 𝐾𝐾 𝑖𝑖𝑖𝑖𝑑𝑑 𝐾𝐾_𝑃𝑃 = 6.55 × 10⁻³ at 725 K, derive the expression for the variation of 𝐾𝐾_𝑃𝑃 𝑇𝑇 with temperature.

∆_𝑟𝑟𝐻𝐻𝐺 𝑇𝑇 = −46.11 × 10⁻³ + �

300

𝑇𝑇 ∆𝐶𝐶𝐺_𝑃𝑃 𝑇𝑇 𝑑𝑑𝑇𝑇

We consider a system consisting of two phases of a pure substance

(1-component) in equilibrium each other. (liquid and gas, for example here) under const.-T and const.-P condition..

The Gibbs energy of this system is given by 𝐺𝐺 = 𝐺𝐺^𝑙𝑙 + 𝐺𝐺^𝑔𝑔

where 𝐺𝐺^𝑙𝑙 and 𝐺𝐺^𝑔𝑔 are the Gibbs energies of the liquid and the gas phase.

Now, if the liquid phase increases with 𝑑𝑑𝑖𝑖^𝑙𝑙 mole (thus, the solid phase decreases with 𝑑𝑑𝑖𝑖^𝑔𝑔 (𝑑𝑑𝑖𝑖^𝑙𝑙 = −𝑑𝑑𝑖𝑖^𝑔𝑔)), the Gibbs energy change is:

𝑑𝑑𝐺𝐺 = 𝑑𝑑𝐺𝐺^𝑙𝑙 + 𝑑𝑑𝐺𝐺^𝑔𝑔 = ^{𝜕𝜕𝐺𝐺}_{𝜕𝜕𝑛𝑛𝑔𝑔}^𝑔𝑔

𝑃𝑃,𝑇𝑇 𝑑𝑑𝑖𝑖^𝑔𝑔 + ^{𝜕𝜕𝐺𝐺}_{𝜕𝜕𝑛𝑛}^𝐶𝐶_𝐶𝐶

𝑃𝑃,𝑇𝑇 𝑑𝑑𝑖𝑖^𝑙𝑙 = ^{𝜕𝜕𝐺𝐺}_{𝜕𝜕𝑛𝑛}^𝑔𝑔_𝑔𝑔

𝑃𝑃,𝑇𝑇 − ^{𝜕𝜕𝐺𝐺}_{𝜕𝜕𝑛𝑛𝐶𝐶}^𝐶𝐶

𝑃𝑃,𝑇𝑇 𝑑𝑑𝑖𝑖^𝑔𝑔 Here, we define chemical potentials, 𝜇𝜇^𝑔𝑔 = ^{𝜕𝜕𝐺𝐺}_{𝜕𝜕𝑛𝑛𝑔𝑔}^𝑔𝑔

𝑃𝑃,𝑇𝑇 and 𝜇𝜇^𝑙𝑙 = ^{𝜕𝜕𝐺𝐺}_{𝜕𝜕𝑛𝑛𝐶𝐶}^𝐶𝐶

𝑃𝑃,𝑇𝑇 , then 𝑑𝑑𝐺𝐺 = 𝜇𝜇^𝑔𝑔 − 𝜇𝜇^𝑙𝑙 𝑑𝑑𝑖𝑖^𝑔𝑔 (constant T and P)

As 𝑑𝑑𝐺𝐺 = 0 is the condition for an equilibrium state, 𝜇𝜇^𝑔𝑔 = 𝜇𝜇^𝑙𝑙 is the condition.

As the condition is cont.-T const.-P, the equilibrium system has the minimum Gibbs energy. This is the ground rule. Here, we consider conditions with which a two-

phase system (liquid & gas here) as the minimum Gibbs energy.

Review of Chapter 3:

Comparison between phase equilibrium and chemical equilibrium

http://cft.fis.uc.pt/eef/FisicaI01/fluids/thermo20.htm (state-1) 300 K, 1 atm

(1)

Phase diagram of water

 The phase diagram indicates that

“at 300 K and 1 atm, the

equilibrium phase of water is liquid water.”

 It means “if water is at 300 K and 1 atm (and with a fixed amount), liquid phase has a smaller Gibbs energy than solid and gas phases.”

 [Quiz] If there is a liquid water in this class room (assuming 300 K), do you think there are some ice (liquid) or vapor water (gas) coexisting?

 Why do you think so and what is the mechanism of it?

𝜕𝜕𝐺𝐺

𝜕𝜕𝑇𝑇 _𝑃𝑃 = −𝑆𝑆 𝜕𝜕𝐺𝐺

𝜕𝜕𝜕𝜕 _𝑇𝑇 = 𝑉𝑉

[J]/[K]=[J/K] [J]/[Pa]=[m³]

Substance �𝑉𝑉 (m³/mol) S^𝐺 (J/mol/K)

H₂O l 0.018E-3 250.5

H₂O g 22.4E-3 188.8

 Typical pressure changes in chemistry are “1 atm → 0.5 atm”, “1atm → 10 atm”, for example, (“1E5 Pa → 0.5E5 Pa”, “1E5 Pa → 1E6 Pa”).

 Typical temperature changes are such as “300 K → 600 K”, etc.

 Thus, the difference between typical pressure and typical temperature (the their changes) are 3-4 orders of magnituede.

 For gas, the difference between �𝑉𝑉 and S^𝐺 are 4 orders of magnitude. Thus, pressure change affect the Gibbs energy of gas in a comparable scale with temperature change.

 For liquid (solid), on the other hand, the difference between �𝑉𝑉 and S^𝐺 are 7 orders of magnitude. Thus, the Gibbs energy is hardly affected by pressure.

Review of Chapter 3:

Comparison between phase equilibrium and chemical equilibrium

H₂O(l) H₂O(l)

H₂O (g) 300 K, V₀ m³

(a Pa)

𝐴𝐴₀ 𝐴𝐴₁

𝜇𝜇^𝑔𝑔 300 𝐾𝐾, 𝑖𝑖

= 𝜇𝜇^𝑙𝑙 (300 𝐾𝐾, 𝑖𝑖) 300 K, V₀ m³

(0 Pa)

H₂O(s)

300 K, V₀ m³ (0 Pa)

𝐴𝐴₂ 𝜇𝜇^𝑙𝑙 300𝐾𝐾, 0𝜕𝜕𝑖𝑖

< 𝜇𝜇^𝑠𝑠 (300𝐾𝐾, 0𝜕𝜕𝑖𝑖)

𝑨𝑨_𝟎𝟎 > 𝑨𝑨_𝟏𝟏 𝑨𝑨_𝟐𝟐 > 𝑨𝑨_𝟎𝟎

Const.-T const.-V condition

H₂O(l)

Initial state

(a non-equilibrium state) State-1

(a non-equilibrium state)

State-2

(the equilibrium state)

𝐺𝐺₀ 𝑮𝑮_𝟐𝟐 > 𝑮𝑮_𝟎𝟎

Const.-T const.-P condition

H₂O (l)

300 K, 3E3 Pa H₂O

(g)

300 K, 1E5 Pa

H₂O (s)

𝐺𝐺₁

𝐺𝐺₂

𝑮𝑮_𝟏𝟏 > 𝑮𝑮_𝟎𝟎

H₂O (g)

H₂O (l)

H₂O (g) H₂O

(l) 𝐺𝐺₃

𝐺𝐺₄ 𝑮𝑮_𝟎𝟎 > 𝑮𝑮_𝟒𝟒

𝑮𝑮_𝟎𝟎 > 𝑮𝑮_𝟑𝟑

 Indeed, 𝑮𝑮_𝟑𝟑 = 𝑮𝑮_𝟒𝟒 if the states where 2 coexisting phases exist are equilibrium states at this const.-T const.-P.

 In this case, 𝜇𝜇^𝑔𝑔 300 𝐾𝐾, 3𝐸𝐸3 𝜕𝜕𝑖𝑖 = 𝜇𝜇^𝑙𝑙 (300 𝐾𝐾, 3𝐸𝐸3 𝜕𝜕𝑖𝑖) H₂O

(l)

Review of Chapter 3:

Comparison between phase equilibrium and chemical equilibrium

 The saturation pressure is the pressure where

chemical potential of liquid and gas phases become identical at a given temperature.

𝜇𝜇^𝑔𝑔 𝑇𝑇 𝐾𝐾, 𝜕𝜕_{𝑠𝑠𝑠𝑠𝑠𝑠} 𝜕𝜕𝑖𝑖 = 𝜇𝜇^𝑙𝑙 (𝑇𝑇 𝐾𝐾, 𝜕𝜕_{𝑠𝑠𝑠𝑠𝑠𝑠} 𝜕𝜕𝑖𝑖)

 For practice, it may also be defined as

𝜇𝜇^𝑔𝑔 𝑇𝑇 𝐾𝐾, 𝜕𝜕_{𝑠𝑠𝑠𝑠𝑠𝑠} 𝜕𝜕𝑖𝑖 = 𝜇𝜇^𝑙𝑙 𝑇𝑇 𝐾𝐾, 1 𝑖𝑖𝑖𝑖𝑚𝑚 ~𝜇𝜇^𝑙𝑙 𝑇𝑇 𝐾𝐾, 𝜕𝜕_{𝑠𝑠𝑠𝑠𝑠𝑠} 𝑖𝑖𝑖𝑖𝑚𝑚

 This equation is more useful when we want to know the saturation pressure of water vapor in a class room, for example.

 Note that (molar) Gibbs energy of water vapor depends on partial pressure of water vapor, not on the total pressure of system. (assuming ideal gas)

<Saturation pressure of water vapor>

So, even when the total pressure of the system is fixed, we may change the Gibbs energy (of the system) by adjusting partial pressures.

 For example, if we initially have 1 mol of H₂O(g), a state where H₂O(g),

H₂(g) and O₂(g) coexist may have the minimum Gibbs energy under const.-P const.-T condition.

0 20000 40000 60000 80000 100000

260 280 300 320

Saturation pressure (Pa)

Temperature (K)

Assuming const.-T condition, and dividing both sides with dP

For reversible processes (thus for thermodynamical equilibrium states) 𝑑𝑑𝐺𝐺 = 𝑉𝑉𝑑𝑑𝜕𝜕 − 𝑆𝑆𝑑𝑑𝑇𝑇

𝜇𝜇 𝑇𝑇, 𝜕𝜕 = 𝜇𝜇𝐺 𝑇𝑇 + 𝑅𝑅𝑇𝑇 ln ⁄𝜕𝜕 𝜕𝜕𝐺

We consider the pressure dependence of gas of pure substance (single- component).

𝜕𝜕𝐺𝐺

𝜕𝜕𝜕𝜕 _𝑇𝑇 = 𝑉𝑉

Then, keeping T and P constant, taking derivative regarding n

𝜕𝜕

𝜕𝜕𝑖𝑖

𝜕𝜕𝐺𝐺

𝜕𝜕𝜕𝜕 _{𝑇𝑇 𝑇𝑇,𝑃𝑃} = 𝜕𝜕𝑉𝑉

𝜕𝜕𝑖𝑖 _{𝑇𝑇,𝑝𝑝}

𝜕𝜕

𝜕𝜕𝑖𝑖

𝜕𝜕𝐺𝐺

𝜕𝜕𝜕𝜕 _{𝑇𝑇 𝑇𝑇,𝑃𝑃} = 𝜕𝜕

𝜕𝜕𝜕𝜕

𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖 _{𝑇𝑇,𝑃𝑃 𝑇𝑇} = 𝜕𝜕𝜇𝜇

𝜕𝜕𝜕𝜕 _𝑇𝑇

Note that up to here ideal gas is not assumed.

Here, we assume ideal gas. Then, using ^{𝜕𝜕𝜕𝜕}

𝜕𝜕𝑛𝑛 𝑇𝑇,𝑝𝑝 = ^{𝑅𝑅𝑇𝑇}_𝑃𝑃

𝜇𝜇 𝑇𝑇, 𝜕𝜕 = 𝜇𝜇𝐺 𝑇𝑇 + 𝑅𝑅𝑇𝑇 ln ⁄𝜕𝜕 𝜕𝜕𝐺

We consider the pressure dependence of gas of pure substance (single- component).

𝜕𝜕𝜇𝜇

𝜕𝜕𝜕𝜕 _𝑇𝑇 = 𝜕𝜕𝑉𝑉

𝜕𝜕𝑖𝑖 _{𝑇𝑇,𝑝𝑝}

𝑑𝑑𝜇𝜇 = 𝑅𝑅𝑇𝑇𝑑𝑑𝜕𝜕

𝜕𝜕 (const.-T)

Taking the integral from the standard pressure (𝜕𝜕𝐺 = 1 𝑏𝑏𝑖𝑖𝑖𝑖) to an arbitrary pressure 𝜕𝜕,

𝜇𝜇 − 𝜇𝜇𝐺 = 𝑅𝑅𝑇𝑇 ln 𝜕𝜕

𝜕𝜕𝐺

Be careful that chemical potential 𝜇𝜇 is molar Gibbs energy for pure substance, but it is not the case when multiple substances are mixed.

3.0. Pressure dependence of chemical potential-

-𝜇𝜇 𝑻𝑻, 𝑷𝑷 = 𝜇𝜇𝐺 𝑻𝑻 + 𝑹𝑹𝑻𝑻 𝒍𝒍𝒍𝒍 𝑷𝑷_𝒋𝒋⁄𝑷𝑷𝐺 for pure substance of an ideal gas -

 These two partial derivative are both definitions of chemical potential, but for different conditions: _{𝜕𝜕𝑛𝑛}^{𝜕𝜕𝐺𝐺}

𝑌𝑌 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝑍𝑍,𝑛𝑛_𝐴𝐴,𝑛𝑛_𝐵𝐵 is for a mixture (A,B,Y,Z) and ^{𝜕𝜕𝐺𝐺}

𝜕𝜕𝑛𝑛 𝑇𝑇,𝑃𝑃 for pure substance.

 For example, ^{𝜕𝜕𝐺𝐺}

𝜕𝜕𝑛𝑛 𝑇𝑇,𝑃𝑃 is the change in Gibbs energy for the pure gas Y when an infinitesimal amount of gas Y (𝑑𝑑𝑖𝑖) is added keeping other parameters constant. In this case, _{𝜕𝜕𝑛𝑛}^{𝜕𝜕𝐺𝐺}

𝑌𝑌 𝑇𝑇,𝑃𝑃 = _𝑛𝑛^𝐺𝐺

𝑌𝑌, because this process only change the amount of Y with fixing other thermodynamical

conditions. Thus, chemical potential is equal to molar Gibbs energy (const.-T and const.-P) for pure substance.

 ^{𝜕𝜕𝐺𝐺}

𝜕𝜕𝑛𝑛_𝑌𝑌 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝑍𝑍,𝑛𝑛_𝐴𝐴,𝑛𝑛_𝐵𝐵 is the change in Gibbs energy for the gas mixture when an infinitesimal amount of gas Y (𝑑𝑑𝑖𝑖_𝑌𝑌) is added keeping other parameters constant. The chemical potential is equal to partial molar Gibbs energy (const.-T and const.-P [total P]) for a mixture.

 These two partial derivative are both definitions of chemical potential, but for different conditions: _{𝜕𝜕𝑛𝑛}^{𝜕𝜕𝐺𝐺}

𝑌𝑌 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝑍𝑍,𝑛𝑛_𝐴𝐴,𝑛𝑛_𝐵𝐵 is for a mixture (A,B,Y,Z) and ^{𝜕𝜕𝐺𝐺}

𝜕𝜕𝑛𝑛 𝑇𝑇,𝑃𝑃 for pure substance.

 (1) For example, if [T= 300 K and P= 1 atm, n=1 mol], “ ^{𝜕𝜕𝐺𝐺}

𝜕𝜕𝑛𝑛 𝑇𝑇,𝑃𝑃 𝑑𝑑𝑖𝑖”

corresponds to the Gibbs energy change by increasing 𝑑𝑑𝑖𝑖 mol of the substance.

 (2) On the other hand, if [T= 300 K, P= 1 atm, n=1 mol],

“ _{𝜕𝜕𝑛𝑛}^{𝜕𝜕𝐺𝐺}

𝑌𝑌 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝑍𝑍,𝑛𝑛_𝐴𝐴,𝑛𝑛_𝐵𝐵 𝑑𝑑𝑖𝑖_𝑌𝑌” corresponds to the Gibbs energy change by

increasing 𝑑𝑑𝑖𝑖_𝑌𝑌 mol of Y. In this case, the partial pressure of the gas is not 1 atm, but _𝑛𝑛 ^{𝑛𝑛𝑌𝑌}

𝑌𝑌+𝑛𝑛_𝑍𝑍+𝑛𝑛_𝐴𝐴+𝑛𝑛_𝐵𝐵 atm. As the Gibbs energy of gas largely depends on the partial pressure, the Gibbs energy change should be different from that of (1).

How can we define Gibbs energy of gas mixture?

3.1. Chemical equilibrium for gases -Difference between

𝝏𝝏𝒍𝒍_𝑨𝑨 𝑻𝑻,𝑷𝑷,𝒍𝒍_𝑩𝑩,𝒍𝒍_𝒀𝒀,𝒍𝒍_𝒁𝒁

and

𝝏𝝏𝒍𝒍 𝑻𝑻,𝑷𝑷-

T₀ K, V₀ m³, P₀ Pa A(g): n_a mol B(g): n_b mol

𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V₀ m³, P_B Pa B(g): n_b mol

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V₀ m³, P_A Pa A(g): n_a mol P𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

The 1^st law: 𝑑𝑑𝑑𝑑 = 𝜕𝜕𝑑𝑑𝑉𝑉 − 𝑇𝑇𝑑𝑑𝑆𝑆

As const.-T const.-V condition, ∆𝑑𝑑 = −𝑇𝑇∆𝑆𝑆

If the gases are ideal gases, because const.-T is assumed, ∆𝑑𝑑 = ∆𝑑𝑑 𝑇𝑇 = 0, hence,

∆𝑆𝑆 = 0.

In addition, for ideal gas, ∆𝐻𝐻 = ∆𝐻𝐻 𝑇𝑇 = 0. Hence, ∆𝐺𝐺 = ∆ 𝐻𝐻 − 𝑇𝑇𝑆𝑆 = 0.

This result means that the Gibbs energy of mixed gas can be defined as the

summation of Gibbs energies of constituent pure gasses holding corresponding partial pressure (for ideal gas).

𝜇𝜇_𝐴𝐴 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖𝐴𝐴 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝐵𝐵 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖_{𝐴𝐴 𝑇𝑇,𝑃𝑃}

𝐴𝐴

= 𝜇𝜇𝐺_𝐴𝐴 𝑇𝑇 + 𝑅𝑅𝑇𝑇 ln 𝜕𝜕_𝐴𝐴⁄𝜕𝜕𝐺

Mixture Pure substance

T₀ K, V₀ m³, P₀ Pa A(g): n_a mol B(g): n_b mol

𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V₀ m³, P_B Pa A(g): n_a mol B(g): n_b mol

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V₀ m³, P_A Pa A(g): n_a mol P𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V₀ m³, P₀ Pa A(g): n_a mol B(g): n_b mol

𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V_B m³, P₀ Pa

B(g): n_b mol 𝑉𝑉_𝐵𝐵 = 𝑉𝑉_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V_A m³, P₀ Pa

A(g): n_a mol P𝑉𝑉_𝐴𝐴 = 𝑉𝑉_0𝑛𝑛 ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

 In this case, we need to consider the

entropy of mixing.

 Indeed, the entropy of mixing is related to entropy change due to volume change.

∆𝐺𝐺 ≠ 0

∆𝑆𝑆 ≠ 0

∆𝐺𝐺 = 0

∆𝑆𝑆 = 0

3.1. Chemical equilibrium for gases

-Gibbs energy in gas mixture (ideal gas)-

and condensed phases (solid, liquid)

Solid Liquid Gas

Effect of pressure on Gibbs energy

 Negligible (< GPa level)

 Negligible (< GPa level)

 Important (comparable with temperature effect)

 Depending on partial pressure If the total

pressure is 1 bar…

 Any solid phase feels 1 bar.

 Any liquid phase feels 1 bar

 Total gas pressure is 1 bar, but partial pressures are

changeable.

*Note there is no “partial pressure” for solid and liquid.

 For pure substance, the chemical potential, which is molar Gibbs energy for pure substance, is as follows, where ”𝐺” denotes the standard state

𝜇𝜇 𝑇𝑇, 𝜕𝜕 = 𝐺𝐺

𝑖𝑖 = 𝜇𝜇𝐺 𝑇𝑇 + 𝑅𝑅𝑇𝑇 ln 𝜕𝜕 𝜕𝜕𝐺⁄

 For example, “𝑅𝑅𝑇𝑇 ln ⁄𝜕𝜕 𝜕𝜕𝐺 ” term at 300 K becomes -1.72 kJ/mol for 0.5x10⁵ Pa, -28.7 kJ/mol for 1 Pa, -57.4 kJ/mol for 1x10^-5 Pa.

 However, the contribution to Gibbs energy of the system (𝜇𝜇𝑖𝑖 , not 𝜇𝜇) is not so large because 𝑖𝑖 is very small for a low pressure.

3.1. Chemical equilibrium for gases

-review-

 The following equation is for pure substance:

𝜇𝜇 𝑇𝑇, 𝜕𝜕 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖 _{𝑇𝑇,𝑃𝑃} = 𝐺𝐺

𝑖𝑖 = 𝜇𝜇𝐺 𝑇𝑇 + 𝑅𝑅𝑇𝑇 ln 𝜕𝜕 𝜕𝜕𝐺⁄ Now we want to know the related equation for mixture.

𝜇𝜇_𝐴𝐴 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖𝐴𝐴 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝐵𝐵,𝑛𝑛_𝑌𝑌,𝑛𝑛_𝑍𝑍

<(system-1) Mixture>

T₀ K, P₀ Pa

A(g): n_A mol, B(g): n_B mol Y(g): n_Y mol, Z(g): n_Z mol

<(system-2) Pure A substance>

T₀ K, P₀ Pa A(g): n_A mol

 For each system, if we add ∆𝑖𝑖_𝐴𝐴 of A with keeping other quantities fixed (T_0, and P₀, and n_B, n_Y and n_Z for system-1), Gibbs energy of the system should be also

changed, ∆𝐺𝐺. However, because the composition of the system different,

∆𝐺𝐺_{𝑠𝑠𝑠𝑠𝑠𝑠−1} ≠ ∆𝐺𝐺_{𝑠𝑠𝑠𝑠𝑠𝑠−2}. And

∆𝑛𝑛lim_𝐴𝐴→0

∆𝐺𝐺_{𝑠𝑠𝑠𝑠𝑠𝑠−1}

∆𝑖𝑖_𝐴𝐴 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖𝐴𝐴 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝐵𝐵,𝑛𝑛_𝑌𝑌,𝑛𝑛_𝑍𝑍

= 𝜇𝜇_𝐴𝐴 ≠ 𝐺𝐺 𝑖𝑖_𝐴𝐴

∆𝑛𝑛lim_𝐴𝐴→0

∆𝐺𝐺_{𝑠𝑠𝑠𝑠𝑠𝑠−2}

∆𝑖𝑖_𝐴𝐴 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖_{𝐴𝐴 𝑇𝑇,𝑃𝑃} = 𝜇𝜇_𝐴𝐴 = 𝐺𝐺 𝑖𝑖_𝐴𝐴

*Note both are definitions of chemical potential, but for different systems.

Mixture Pure substance

be replaced with the chemical potential for a pure gas substance with the corresponding partial pressure.

T₀ K, V₀ m³, P₀ Pa A(g): n_A mol B(g): n_B mol

𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_0𝑛𝑛 ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

where 𝐺𝐺𝑠𝑠𝑠𝑠𝑠𝑠−𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐺𝐺𝑝𝑝𝑝𝑝𝑟𝑟𝑝𝑝−𝐴𝐴 + 𝐺𝐺𝑝𝑝𝑝𝑝𝑟𝑟𝑝𝑝−𝐵𝐵 [for ideal gas].

𝜇𝜇_𝐴𝐴 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖𝐴𝐴 𝑇𝑇,𝑃𝑃,𝑛𝑛_𝐵𝐵 = 𝜕𝜕𝐺𝐺

𝜕𝜕𝑖𝑖_{𝐴𝐴 𝑇𝑇,𝑃𝑃}

𝐴𝐴

= 𝜇𝜇𝐺_𝐴𝐴 𝑇𝑇 + 𝑅𝑅𝑇𝑇 ln 𝜕𝜕_𝐴𝐴⁄𝜕𝜕𝐺

T₀ K, V₀ m³, P_B Pa B(g): n_B mol

𝜕𝜕_𝐵𝐵 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐵𝐵}

𝐴𝐴+𝑛𝑛_𝐵𝐵

T₀ K, V₀ m³, P_A Pa A(g): n_A mol P𝜕𝜕_𝐴𝐴 = 𝜕𝜕_{0 𝑛𝑛} ^{𝑛𝑛𝐴𝐴}

𝐴𝐴+𝑛𝑛_𝐵𝐵

<System of mixture> <System of pure A > <System of pure B >

 This idea can be extended to more than binary systems, then we can evaluate the pressure dependence of chemical potential of mixture as:

3.1. Chemical equilibrium for gases

-review-

 Using the obtained pressure dependence of Gibbs energy of gas phase in a

mixture, we can further write down “dG = ∑_𝑚𝑚𝜇𝜇_𝑚𝑚𝑑𝑑𝑖𝑖_𝑚𝑚” to find dG = 0 condition, which is the condition for equilibrium state.

 Then, as a result, following chemical equilibrium equations are obtained.

 Again, please recall that these equations give us the condition where the Gibbs energy of the system is minimized. So, we can get the same result if we directly evaluate the Gibbs energy of the system and find the condition with which it is minimized.

∆_𝑟𝑟𝐺𝐺𝐺 = −𝑅𝑅𝑇𝑇 ln 𝐾𝐾_𝑃𝑃(𝑇𝑇)

𝐾𝐾_𝑃𝑃 𝑇𝑇 = 𝑄𝑄_{𝑝𝑝𝑒𝑒} = 𝜕𝜕_𝑌𝑌⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑌𝑌} 𝜕𝜕_𝑍𝑍⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑍𝑍}

𝜕𝜕_𝐴𝐴⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐴𝐴} 𝜕𝜕_𝐵𝐵⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐵𝐵}

𝑝𝑝𝑒𝑒

= 𝜕𝜕_𝑌𝑌^{𝜈𝜈𝑌𝑌}𝜕𝜕_𝑍𝑍^{𝜈𝜈𝑍𝑍}

𝜕𝜕_𝐴𝐴^{𝜈𝜈𝐴𝐴}𝜕𝜕_𝐵𝐵^{𝜈𝜈𝐵𝐵} × 𝜕𝜕𝐺 ^{𝜈𝜈𝐴𝐴+𝜈𝜈𝐵𝐵−𝜈𝜈𝐵𝐵−𝜈𝜈𝐵𝐵}

𝑝𝑝𝑒𝑒

*the subscript eq emphasizes that the partial pressures are in an equilibrium.

𝜕𝜕𝐺 = 1 𝑏𝑏𝑖𝑖𝑖𝑖

∆_𝑟𝑟𝐺𝐺𝐺 = 𝜈𝜈_𝑌𝑌𝜇𝜇𝐺_𝑌𝑌 𝑇𝑇 + 𝜈𝜈_𝑍𝑍𝜇𝜇𝐺_𝑍𝑍 𝑇𝑇 − 𝜈𝜈_𝐴𝐴𝜇𝜇𝐺_𝐴𝐴(𝑇𝑇) − 𝜈𝜈_𝐵𝐵𝜇𝜇𝐺_𝐵𝐵(𝑇𝑇)

𝐾𝐾_𝑃𝑃 𝑇𝑇 = 𝑖𝑖𝑒𝑒𝑒𝑒 −∆_𝑟𝑟𝐺𝐺𝐺 𝑅𝑅𝑇𝑇

For “𝜈𝜈_𝐴𝐴A(g) + 𝜈𝜈_𝐵𝐵B(g) ⇌ 𝜈𝜈_𝑌𝑌Y(g) + 𝜈𝜈_𝑍𝑍Z(g)”

reaction at 1000 K: H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)

(Quiz-2) Suppose that we have a mixture of the gases H₂(g), CO₂(g), CO(g) and H₂O(g) at 1000 K at 1 bar, with 𝜕𝜕_𝐻𝐻2 = 0.200 𝑏𝑏𝑖𝑖𝑖𝑖, 𝜕𝜕_{𝐶𝐶𝐶𝐶2} = 0.200 𝑏𝑏𝑖𝑖𝑖𝑖,

𝜕𝜕_{𝐶𝐶𝐶𝐶} = 0.300 𝑏𝑏𝑖𝑖𝑖𝑖, and 𝜕𝜕_{𝐻𝐻2𝐶𝐶} = 0.300 𝑏𝑏𝑖𝑖𝑖𝑖. Is the reaction described by the

equation "H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)" at equilibrium? If not, in what direction (left or right) will the reaction proceed to attain equilibrium for

const.-T cons.t-P condition?

(Quiz-3) Please determine the equilibrium partial pressures at 1000 K at 1 bar.

∆_𝒇𝒇𝑮𝑮𝐺 (kJ/mol) @ 1000 K

H₂(g) 0

CO₂(g) -394.9

CO(g) -155.4

H₂O(g) -192.6

Quiz-answer

(Quiz-1) Please determine the equilibrium constant for the following reaction at 1000 K: H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)

∆_𝑟𝑟𝐺𝐺𝐺 = 𝜈𝜈_𝑌𝑌𝜇𝜇𝐺_𝑌𝑌 𝑇𝑇 + 𝜈𝜈_𝑍𝑍𝜇𝜇𝐺_𝑍𝑍 𝑇𝑇 − 𝜈𝜈_𝐴𝐴𝜇𝜇𝐺_𝐴𝐴 𝑇𝑇 − 𝜈𝜈_𝐵𝐵𝜇𝜇𝐺_𝐵𝐵 𝑇𝑇

= 𝜇𝜇𝐺_{𝐶𝐶𝐶𝐶} 𝑇𝑇 + 𝜇𝜇𝐺_{𝐻𝐻2𝐶𝐶} 𝑇𝑇 − 𝜇𝜇𝐺_𝐻𝐻2 𝑇𝑇 − 𝜇𝜇𝐺_{𝐶𝐶𝐶𝐶2} 𝑇𝑇

= −200.3 + −192.6 − 0 − 395.9 = 3 kJ/mol 𝐾𝐾_𝑃𝑃 𝑇𝑇 = 𝑖𝑖𝑒𝑒𝑒𝑒 −∆_𝑟𝑟𝐺𝐺𝐺

𝑅𝑅𝑇𝑇 = 𝑖𝑖𝑒𝑒𝑒𝑒 −

3 × 10³

8.31 × 1000 = 0.697

∆_𝒇𝒇𝑮𝑮𝐺 (kJ/mol) @ 1000 K

H₂(g) 0

CO₂(g) -395.9

CO(g) -200.3

H₂O(g) -192.6

H₂O(g) at 1000 K at 1 bar, with 𝜕𝜕_𝐻𝐻2 = 0.200 𝑏𝑏𝑖𝑖𝑖𝑖, 𝜕𝜕_{𝐶𝐶𝐶𝐶2} = 0.200 𝑏𝑏𝑖𝑖𝑖𝑖, 𝜕𝜕_{𝐶𝐶𝐶𝐶} = 0.300 𝑏𝑏𝑖𝑖𝑖𝑖, and 𝜕𝜕_{𝐻𝐻2𝐶𝐶} = 0.300 𝑏𝑏𝑖𝑖𝑖𝑖. Is the reaction described by the equation "H₂(g) + CO₂(g) ⇌

CO(g) + H₂O(g)" at equilibrium? If not, in what direction (left or right) will the reaction proceed to attain equilibrium for const.-T cons.t-P condition?

𝐾𝐾_𝑃𝑃 𝑇𝑇 = 𝑄𝑄_{𝑝𝑝𝑒𝑒} = 𝜕𝜕_𝑌𝑌⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑌𝑌} 𝜕𝜕_𝑍𝑍⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑍𝑍}

𝜕𝜕_𝐴𝐴⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐴𝐴} 𝜕𝜕_𝐵𝐵⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐵𝐵}

𝑝𝑝𝑒𝑒

= 𝜕𝜕_𝑌𝑌^{𝜈𝜈𝑌𝑌}𝜕𝜕_𝑍𝑍^{𝜈𝜈𝑍𝑍}

𝜕𝜕_𝐴𝐴^{𝜈𝜈𝐴𝐴}𝜕𝜕_𝐵𝐵^{𝜈𝜈𝐵𝐵} × 𝜕𝜕𝐺 ^{𝜈𝜈𝐴𝐴+𝜈𝜈𝐵𝐵−𝜈𝜈𝐵𝐵−𝜈𝜈𝐵𝐵}

𝑝𝑝𝑒𝑒

We need to use an equation like:

First, we evaluate “reaction quotient” using the following equation:

𝑄𝑄 = 𝜕𝜕_𝑌𝑌⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑌𝑌} 𝜕𝜕_𝑍𝑍⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑍𝑍}

𝜕𝜕_𝐴𝐴⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐴𝐴} 𝜕𝜕_𝐵𝐵⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐵𝐵} = 0.300 × 0.300

0.200 × 0.200 = 2.25

Because this value is larger than 𝐾𝐾_𝑃𝑃(= 0.697), the system should evolve to decrease 𝑄𝑄 (non-equilibrium state now). Thus, partial pressure of reactants will increase, which means the reaction will proceed from the right to the left.

*Be careful that 𝑄𝑄 is not identical to 𝑄𝑄_{𝑝𝑝𝑒𝑒}. 𝑄𝑄 = 𝑄𝑄_{𝑝𝑝𝑒𝑒} = 𝐾𝐾_𝑃𝑃 𝑇𝑇 is achieved when the system (thus partial pressures) are of equilibrium state.

Quiz answer

(Quiz-3) Please determine the equilibrium partial pressures at 1000 K at 1 bar.

𝐾𝐾_𝑃𝑃 𝑇𝑇 = 𝑄𝑄_{𝑝𝑝𝑒𝑒} = 𝜕𝜕_𝑌𝑌⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑌𝑌} 𝜕𝜕_𝑍𝑍⁄𝜕𝜕𝐺 ^{𝜈𝜈𝑍𝑍}

𝜕𝜕_𝐴𝐴⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐴𝐴} 𝜕𝜕_𝐵𝐵⁄𝜕𝜕𝐺 ^{𝜈𝜈𝐵𝐵}

𝑝𝑝𝑒𝑒

= (0.300 + 𝜉𝜉_{𝑝𝑝𝑒𝑒})(0.300 + 𝜉𝜉_{𝑝𝑝𝑒𝑒})

(0.200 − 𝜉𝜉_{𝑝𝑝𝑒𝑒})(0.200 − 𝜉𝜉_{𝑝𝑝𝑒𝑒}) _{𝑝𝑝𝑒𝑒} = 0.697 If we define the extent of reaction with 𝜉𝜉, the partial pressures at

equilibrium state is given as follows (𝜉𝜉 becomes 𝜉𝜉_{𝑝𝑝𝑒𝑒}):

𝜕𝜕_𝐻𝐻2 = (0.200 − 𝜉𝜉_{𝑝𝑝𝑒𝑒}) 𝑏𝑏𝑖𝑖𝑖𝑖 𝜕𝜕_{𝐶𝐶𝐶𝐶2} = (0.200 − 𝜉𝜉_{𝑝𝑝𝑒𝑒}) 𝑏𝑏𝑖𝑖𝑖𝑖

𝜕𝜕_{𝐶𝐶𝐶𝐶} = (0.300 + 𝜉𝜉_{𝑝𝑝𝑒𝑒}) 𝑏𝑏𝑖𝑖𝑖𝑖 𝜕𝜕_{𝐻𝐻2𝐶𝐶} = (0.300 + 𝜉𝜉_{𝑝𝑝𝑒𝑒}) 𝑏𝑏𝑖𝑖𝑖𝑖 Then,

0.300 + 𝜉𝜉_{𝑝𝑝𝑒𝑒} = 0.697 0.200 − 𝜉𝜉_{𝑝𝑝𝑒𝑒} 𝜉𝜉_{𝑝𝑝𝑒𝑒} = −0.0725

𝜕𝜕_𝐻𝐻2 = 𝜕𝜕_{𝐶𝐶𝐶𝐶2} = 0.273 𝑏𝑏𝑖𝑖𝑖𝑖 𝜕𝜕_{𝐶𝐶𝐶𝐶} = 𝜕𝜕_{𝐻𝐻2𝐶𝐶} = 0.227 𝑏𝑏𝑖𝑖𝑖𝑖

Applying this value to the partial pressure expressions, we can obtain the following equilibrium partial pressures.

(Problem) Initially, we have 1 mol of CH₄(g) and 1 mol of H₂O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH₄(g), H₂O(g), CO(g), H₂(g), C₂H₂(g). Please determine the equilibrium amounts of these species.

3.1. Chemical equilibrium for gases

-an example-

(Problem) Initially, we have 1 mol of CH₄(g) and 1 mol of H₂O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH₄(g), H₂O(g), CO(g), H₂(g), C₂H₂(g). Please determine the equilibrium amounts of these species.

CH₄(g) + H₂O(g) →

n_CH4 CH₄(g) + n_H2O H₂O(g) + n_CO CO(g) + n_H2 H₂(g) + n_C2H2 C₂H₂(g)

 For mass conservation, we have 3 conditions.

 [H] 4+2=4 n_CH4 + 2 n_H2O +2 n_H2 +2 n_C2H2

 [C] 1= n_CH4 + n_CO + 2n_C2H2

 [O] 1= n_H2O + n_CO

 We need 2 additional equations to determine 5 parameters. At equilibrium states, any balance reactions should be of equilibrium states. Then, we consider 2 of them.

 [eq.1] CH₄(g) + H₂O(g) ⇌ 3H2(g) + CO(g)

 [eq.2] CH₄(g) + CO(g) ⇌ C2H₂(g) + H₂O(g)

 Now we have 5 equations for 5 unknown variables, then we can determine the 5 variables.

(Problem) Initially, we have 1 mol of CH₄(g) and 1 mol of H₂O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH₄(g), H₂O(g), CO(g), H₂(g), C₂H₂(g). Please determine the equilibrium amounts of these species.

[eq.1] CH₄(g) + H₂O(g) ⇌ 3H2(g) + CO(g)

[eq.2] CH₄(g) + CO(g) ⇌ C2H₂(g) + H₂O(g) 𝐾𝐾_𝑃𝑃,1 𝑇𝑇 = 𝑖𝑖𝑒𝑒𝑒𝑒 −∆_𝑟𝑟𝐺𝐺𝐺₁

𝑅𝑅𝑇𝑇 = 𝜕𝜕_𝐻𝐻2³ 𝜕𝜕_{𝐶𝐶𝐶𝐶}

𝜕𝜕_{𝐶𝐶𝐻𝐻4}𝜕𝜕_{𝐻𝐻2𝐶𝐶}

𝐾𝐾_𝑃𝑃,2 𝑇𝑇 = 𝑖𝑖𝑒𝑒𝑒𝑒 −∆_𝑟𝑟𝐺𝐺𝐺₂

𝑅𝑅𝑇𝑇 = 𝜕𝜕_{𝐶𝐶2𝐻𝐻2}𝜕𝜕_{𝐻𝐻2𝐶𝐶}

𝜕𝜕_{𝐶𝐶𝐻𝐻4}𝜕𝜕_{𝐶𝐶𝐶𝐶}

 For example, if the balance reaction given in [eq.1] is not of equilibrium state, some reaction (to right or to left) should happen to decrease the Gibbs energy of the system.

 Hence, any kind of balance reaction between constituent gases should achieve equilibrium conditions.

where partial pressures are proportional to molar fractions.

Quiz

(Quiz-1) Please determine the equilibrium constant for the following reaction at 1000 K: H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)

(Quiz-2) Suppose that we have a mixture of the gases H₂(g), CO₂(g), CO(g) and H₂O(g) at 1000 K at 1 bar, with 𝜕𝜕_𝐻𝐻2 = 0.200 𝑏𝑏𝑖𝑖𝑖𝑖, 𝜕𝜕_{𝐶𝐶𝐶𝐶2} = 0.200 𝑏𝑏𝑖𝑖𝑖𝑖, 𝜕𝜕_{𝐶𝐶𝐶𝐶} = 0.300 𝑏𝑏𝑖𝑖𝑖𝑖, and 𝜕𝜕_{𝐻𝐻2𝐶𝐶} = 0.300 𝑏𝑏𝑖𝑖𝑖𝑖. Is the reaction described by the equation

"H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)" at equilibrium? If not, in what direction (left or right) will the reaction proceed to attain equilibrium for const.-T cons.t-P condition?

(Quiz-3) Please determine the equilibrium partial pressures at 1000 K at 1 bar.

∆_𝒇𝒇𝑮𝑮𝐺 (kJ/mol) @ 1000 K

H₂(g) 0

CO₂(g) -394.9

CO(g) -155.4

H₂O(g) -192.6

(Problem) Initially, we have 1 mol of CH₄(g) and 1 mol of H₂O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH₄(g), H₂O(g), CO(g), H₂(g), C₂H₂(g). Please determine the equilibrium amounts of these species.